Proof of thermodynamic stability condition

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SUMMARY

The forum discussion centers on the proof of the thermodynamic stability condition as presented in Kardar's Statistical Mechanics course. The key focus is on the relationship between intensive and extensive variables in a homogeneous system at equilibrium, specifically how variations in energy (E), volume (x), and particle number (N) affect the intensive variables (temperature T, current density J, and chemical potential μ) in two subsystems (A and B). The conclusion reached is that, under equilibrium conditions, the intensive variables of both subsystems must be equal, leading to the relationships δT_A = -δT_B, δJ_A = -δJ_B, and δμ_A = -δμ_B.

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chimay
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Question about a mathematical detail in the proof of the thermodynamic stability relation
I am watching Kardar's Statistical Mechanics course in my spare time and I am struggling to understand a mathematical detail in the proof of the thermodynamic stability condition. See Eq. I.62 here.

The author considers a homogeneous system at equilibrium with intensive and extensive variables (T,J,\mu) and (E,x,N), respectively. Then, He imagines that the latter system is divided into two subsystems (A and B) that can exhange energy. Under the assumption that E,x and N are conserved, we have \delta E_A=-\delta E_B, \delta x_A=-\delta x_B and \delta N_A=-\delta N_B. What is not clear to me is the statement "Since the intensive variables are themselves functions of the extensive coordinates, to first order in the variations of (E, x, N), we have \delta T_A=-\delta T_B, \delta J_A=-\delta J_B and \delta \mu_A=-\delta \mu_B."

Can anyone explain to me the previous statement more in detail?

Thank you!
 
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If, for example, ##T=f(E)##, then to the first order, ##\delta T=f'\, \delta E##.
Then, ##\delta E_A = -\delta E_B## implies ##\delta T_A = -\delta T_B##.
 
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Hi Hill, thank you for your reply.
I understand your point:
\delta T_A = \frac{\partial f}{\partial E}\bigg\rvert_{T_A} \delta E_A and
\delta T_B = \frac{\partial f}{\partial E}\bigg\rvert_{T_B} \delta E_B.

However, since the two subsystems are not necessarily equal, even though \delta E_A = - \delta E_B, \delta T_A \ne \delta T_B because \frac{\partial f}{\partial E}\bigg\rvert_{T_A} \ne \frac{\partial f}{\partial E}\bigg\rvert_{T_B}.

Where am I wrong?
 
chimay said:
However, since the two subsystems are not necessarily equal, even though δEA=−δEB, δTA≠δTB because ∂f∂E|TA≠∂f∂E|TB.
They say, "The two subsystems, A and B, initially have the same values for the intensive variables". That is, ##T_A=T_B##. Then, ##\frac{\partial f}{\partial E}\bigg\rvert_{T_A} = \frac{\partial f}{\partial E}\bigg\rvert_{T_B}##.
 
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You are right! I read it so many times but I missed that. Indeed, since the overall system is at equilibrium, it makes sense that the two "subsystems" have the same values for the intensive variables. Kardar's book is very dense of information and I find it hard to follow sometimes...

Thank you for your help!
 
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